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Unformatted text preview: Math 25: Advanced Calculus UC Davis, Spring 2011 Math 25 Solutions to Homework Assignment #6 1. Let sequences ( s n ) n =1 and ( t n ) n =1 converge to respective limits S and T . Fix some > 0. By definition, there exist numbers m s ,m t such that for any n s m s and n t m t ,  s n s S  < 2 and  t n t T  < 2 . If n max { m s ,m t } , by the triangle inequality,  ( s n t n ) ( S T )  =  ( s n S ) ( t n T )   s n S  +  t n T  < 2 + 2 = . By definition, then, s n t n S T as n . 2. For ( s n ) n =1 and ( t n ) n =1 such that s n = n + 1 , t n = n, Both sequences diverge to infinity, and hence the reasoning in problem 1. does not apply. However, for any n , since a b = ( a b )( a + b ), s n t n = n + 1 n = ( n + 1) n n + 1 + n = 1 n + 1 + n Now, since 1 n + 1 + n 1 n And because 1 n 0 as n , by the squeeze theorem, ( s n t n ) n...
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This note was uploaded on 10/16/2011 for the course MATH 34 taught by Professor Wiley during the Winter '11 term at UC Merced.
 Winter '11
 Wiley
 Differential Equations, Calculus, Equations, Limits

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